Polylogarithm ladders for the tribonacci and n-nacci constants

Solution 1:

I'll change $x \rightarrow 1+x$ to have \begin{equation} \left(1+x\right)^n\left(1-x\right) = 1,\quad 0 < x < 1 \label{eq1} \tag{1} \end{equation} Consider \begin{align} f_n &= 4\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^2}\right) + \operatorname{Li}_2\left(\dfrac{1+x}{\left(1+x\right)^n}\right) - 2\operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^n}\right) - \phantom{a} \\ &- 2\operatorname{Li}_2\left(-1-x\right) \end{align} where $x$ is the solution to $(1)$ with parameter $n$.

I restate here main functional equations for the dilogarithm (most of them can be found here)

\begin{align} \operatorname{Li}_2(x)+\operatorname{Li}_2(-x) &= \dfrac{1}{2}\operatorname{Li}_2(x^2) \label{dl2} \tag{2} \\ \operatorname{Li}_2\left(1-x\right)+\operatorname{Li}_2\left(1-\dfrac{1}{x}\right) &= -\dfrac{1}{2}\ln^2 x \label{dl3} \tag{3} \\ \operatorname{Li}_2(x)+\operatorname{Li}_2\left(1-x\right) &= \dfrac{1}{6}\pi^2 - \ln\left(1-x\right)\ln x \label{dl4} \tag{4} \\ \operatorname{Li}_2(-x)-\operatorname{Li}_2\left(1-x\right)+\dfrac{1}{2}\operatorname{Li}_2\left(1-x^2\right) &= -\dfrac{1}{12}\pi^2 - \ln\left(1+x\right)\ln x \label{dl5} \tag{5} \\ \operatorname{Li}_2(-x)+\operatorname{Li}_2\left(-\dfrac{1}{x}\right) &= -\dfrac{1}{6}\pi^2 - \dfrac{1}{2}\ln^2 x \label{dl6} \tag{6} \end{align}

Now use $\left(1+x\right)^n\left(1-x\right) = 1$ to get rid of $n$-th power of $(1+x)$ in $f_n$ and start to use dilogarithm identities \begin{align*} f_n &\overset{(1)}{=} 4\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{red}{\operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^2}\right)} + \operatorname{Li}_2\left(1-x^2\right) - 2\operatorname{Li}_2\left(1-x\right) - 2\operatorname{Li}_2\left(-1-x\right) \\ &\overset{(2)}{=} \color{red}{2}\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{red}{2\operatorname{Li}_2\left(-\dfrac{1}{1+x}\right)} + \color{blue}{\operatorname{Li}_2\left(1-x^2\right)} - \color{blue}{2\operatorname{Li}_2\left(1-x\right)} - 2\operatorname{Li}_2\left(-1-x\right) \\ &\overset{(5)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{magenta}{2\operatorname{Li}_2\left(-\dfrac{1}{1+x}\right)} - \color{blue}{\dfrac{1}{6}\pi^2} - \color{blue}{2\ln\left(1+x\right)\ln x} - \color{blue}{2\operatorname{Li}_2(-x)}- \color{magenta}{2\operatorname{Li}_2\left(-1-x\right)} \\ &\overset{(6)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) + \color{magenta}{\dfrac{1}{3}\pi^2} + \color{magenta}{\ln^2\left(1+x\right)} - \dfrac{1}{6}\pi^2 - 2\ln\left(1+x\right)\ln x - \color{green}{2\operatorname{Li}_2(-x)} \\ &\overset{(3)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) + \dfrac{1}{3}\pi^2 + \ln^2\left(1+x\right) - \dfrac{1}{6}\pi^2 - 2\ln\left(1+x\right)\ln x + \color{green}{\ln^2\left(1+x\right)} + \phantom{a} \\ &+ \color{green}{2\operatorname{Li}_2\left(\dfrac{x}{1+x}\right)} \\ &= \color{red}{2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right)} + \color{red}{2\operatorname{Li}_2\left(\dfrac{x}{1+x}\right)} + \dfrac{1}{6}\pi^2 + 2\ln^2\left(1+x\right)- 2\ln\left(1+x\right)\ln x \\ &\overset{(4)}{=} \color{red}{\dfrac{1}{3}\pi^2} - \color{red}{2\ln\left(\dfrac{1}{1+x}\right)\ln\left(\dfrac{x}{1+x}\right)} + \dfrac{1}{6}\pi^2 + 2\ln^2\left(1+x\right)- 2\ln\left(1+x\right)\ln x \\ &= \dfrac{\pi^2}{2} \end{align*}